Problem 78 - University Calculus: Early Transcendentals

Question

Find \(y^{\prime \prime}\) in Exercises \(71-78\). $$y=\sin \left(x^{2} e^{x}\right)$$

Step-by-Step Solution

Answer

The second derivative is \( y'' = -\sin(x^2 e^x) (e^x(x^2 + 2x))^2 + \cos(x^2 e^x) e^x(x^2 + 4x + 2) \).

Step 1: Identify the Function

The function given is \( y = \sin(x^2 e^x) \). This involves a composition of functions, specifically a sine function and a product inside it. Our goal is to find the second derivative of this function.

Step 2: Use the Chain Rule for the First Derivative

To find the first derivative \( y' \), use the chain rule:\[ y' = \cos(x^2 e^x) \cdot \frac{d}{dx}(x^2 e^x) \]First, compute the derivative of the inner function \( u = x^2 e^x \).

Step 3: Differentiate the Inner Function

Differentiate \( u = x^2 e^x \) using the product rule:\[ \frac{d}{dx}(x^2 e^x) = x^2 \cdot \frac{d}{dx}(e^x) + e^x \cdot \frac{d}{dx}(x^2) = x^2 e^x + 2x e^x \]Thus, \( \frac{d}{dx}(x^2 e^x) = e^x(x^2 + 2x) \).

Step 4: Compute the First Derivative

Substitute back into the expression for \( y' \):\[ y' = \cos(x^2 e^x) \cdot e^x(x^2 + 2x) \]

Step 5: Differentiate Once More for the Second Derivative

Apply the product rule to find \( y'' \). Take the derivative of the product:\[ y'' = \frac{d}{dx}(\cos(x^2 e^x)) \cdot e^x(x^2 + 2x) + \cos(x^2 e^x) \cdot \frac{d}{dx}(e^x(x^2 + 2x)) \]Evaluate each part separately.

Step 6: Differentiate the Cosine Term

Using the chain rule, \[ \frac{d}{dx}(\cos(x^2 e^x)) = -\sin(x^2 e^x) \cdot \frac{d}{dx}(x^2 e^x) = -\sin(x^2 e^x) \cdot e^x(x^2 + 2x) \]

Step 7: Differentiate the Second Product

Now differentiate \( e^x(x^2 + 2x) \) using the product rule:\[ \frac{d}{dx}(e^x(x^2 + 2x)) = e^x \cdot \frac{d}{dx}(x^2 + 2x) + (x^2 + 2x) \cdot \frac{d}{dx}(e^x) \]\[ = e^x(2x + 2) + (x^2 + 2x)e^x = e^x(2x + 2 + x^2 + 2x) = e^x(x^2 + 4x + 2) \]

Step 8: Combine Results for Second Derivative

Substitute the results into the expression for \( y'' \):\[ y'' = -\sin(x^2 e^x) \cdot (e^x(x^2 + 2x))^2 + \cos(x^2 e^x) \cdot e^x(x^2 + 4x + 2) \]

Key Concepts

Chain Rule

When dealing with differentiation, the chain rule is an invaluable tool that allows us to tackle complex functions involving compositions. Simply put, the chain rule helps us find the derivative of a composite function by differentiating the outer function and then multiplying it by the derivative of the inner function. Let's break it down step-by-step using the function from our example,

• Identify the outer and inner functions: In the function \( y = \sin(x^2 e^x) \), the outer function is \( \sin(u) \) with \( u = x^2 e^x \) as the inner function.
• Apply the derivative to the outer function and multiply by the derivative of the inner function: For our example, the derivative of \( \sin(u) \) is \( \cos(u) \cdot u' \).

This gives us \( y' = \cos(x^2 e^x) \cdot \frac{d}{dx}(x^2 e^x) \). Turning to our inner function, the next step involves using the product rule.

Product Rule

In calculus, the product rule helps us differentiate functions that are products of two or more functions. When a function is expressed as a product, say \( uv \), the derivative \( \frac{d}{dx}(uv) \) is computed as \( u'v + uv' \). In our example, the inner function is a product \( x^2 e^x \). Here's how you apply the product rule efficiently:

• Differentiate each part: For \( x^2 e^x \), let \( u = x^2 \) and \( v = e^x \).
• Apply the product rule: \( \frac{d}{dx}(x^2 e^x) = \frac{d}{dx}(x^2) \cdot e^x + x^2 \cdot \frac{d}{dx}(e^x) \).
• Calculate the derivatives: \( \frac{d}{dx}(x^2) = 2x \) and \( \frac{d}{dx}(e^x) = e^x \).

This results in \( x^2 e^x + 2x e^x = e^x(x^2 + 2x) \). With this, we can now continue finding higher derivatives using these techniques.

Differentiation Techniques

Differentiation is a key skill in calculus, providing a way to find rates of change and slopes of curves. Several techniques aid in finding derivatives, including the chain and product rules already discussed. Utilizing these rules in tandem allows us to handle more complex derivatives efficiently. Here's a strategic approach to finding a second derivative:

• First Derivative: Begin by applying relevant rules to the original function. For our example, the first derivative, \( y' = \cos(x^2 e^x) \cdot e^x(x^2 + 2x) \), combines chain and product rules.
• Second Derivative: Apply the product rule again to \( y' \) to tackle \( y'' \). This requires differentiating each term of the form \( f \cdot g \) according to \( f'g + fg' \).
• Integration of Techniques: Don't forget to revisit previously used rules like the chain rule when differentiating composite functions again, just as in the cosine term for this example.

Understanding and mastering these techniques simplifies differentiation, allowing us to solve complex problems in calculus with greater confidence and accuracy.